College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 33



Work Step by Step

$$e^{1-5x}=793$$ Here we would take the logarithm of both sides. Since the left side involves $e$, we would take the natural logarithm. $$\ln(e^{1-5x})=\ln793$$ $$(1-5x)\ln e=\ln793$$ $$1-5x=\ln793$$ (for $\ln e=1$) $$5x=1-\ln793$$ $$x=\frac{1-\ln793}{5}$$ $$x\approx-1.14$$
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